IOQM mock test Narayana 12th July

Q1. Two friends, Marco and Ian, are talking about their ages. Ian says, "My age is a zero of a polynomial with integer coefficients." Having seen the polynomial p(x) Ian was talking about, Marco exclaims,

"You mean, you are seven years old? Oops, sorry I miscalculated! p(7) = 77 and not zero."

"Yes, I am older than that," Ian's agreeing reply. Then Marco mentioned a certain number, but realizes after a while that he was wrong again because the value of the polynomial at that number is 85. Ian sighs, "I am even older than that number." Determine Ian's age.

S1.
Hint: (a-b) divides (a^n - b^n) for all integer 'n'.
Let the root be 'a'.
p(a) = 0
p(7) = 77
p(b) = 85 and b > 7.
a > b > 7.
For a polynomial with integer coefficients and x,y as integers:
x-y | p(x) - p(y)

So a-7 | p(a) - p(7)
a-7 | -77
So a-7 can be 1,7,11,77
=> a can be {8,14,18,84}

b - 7 | 8
=> b-7 can be 1,2,4,8
=> b can be {8,9,11,15}

a-b| 85
=> a-b can be 1,5,17,85

Only option which works is a = 14, b = 9
=> a = 14 = answer.

Q2. Find the number of triples of positive integers (m,p,q) such that

2^m.p^2 + 27 = q^3, and p is a prime.
S2.
2^m.p^2 + 27 = q^3
=> 2^m.p^2 = q^3 - 27 = (q-3)(q^2 + 3q + 9)
LHS is even. (q^2 + 3q + 9) is odd. => q-3 is even => q is odd.
And (q-3) will contain 2^m since other term is odd.
And p^2 will be divided among the 2 terms.

Case 1:
q-3 = 2^m.p^2 and q^2 + 3q + 9 = 1 which is not possible since q^2 + 3q + 9 > 1 for each positive q.

Case 2:
q-3 = 2^m.p and q^2 + 3q + 9 = p
q - 3 = 2^m.(q^2 + 3q + 9)
This is clearly impossible since RHS is much much larger than LHS.

Case 3:
q-3 = 2^m and q^2 + 3q + 9 = p^2
q = 2^m + 3
(2^m + 3)^2 + 3(2^m + 3) + 9 = p^2
2^(2m) + 9 + 6.2^m + 3.2^m + 18 = p^2
Let 2^m = x
x^2 + 27 + 9x = p^2
x^2 + 2.9/2.x + 81/4 - 81/4 + 27 = p^2
(x + 9/2)^2 + 27/4 = p^2
(2x + 9)^2 + 27 = 4p^2
(2p + 2x + 9) (2p - 2x - 9) = 27

Case 3.1
2p + 2x + 9 = 27
2p - 2x - 9 = 1
p = 7, 4x + 18 = 26 4x = 8, x = 2 => m = 1 since x = 2^m

Case 3.2
2p + 2x + 9 = 9
2p - 2x - 9 = 3
4p = 12, p = 3, 4x + 18 = 6 doesn't work.

So p = 7, m = 1 q-3 = 2 => q = 5 which satisfies 25 + 15 + 9 = 49
Only triplet is (m,p,q) = (1,7,5) 



3. The quadratic polynomial (f(x)) has a zero at (x=2). The polynomial (f(f(x))) has only one real zero, at (x=5). Compute


[

|18f(0)|.

]


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4.  A rational number written in base eight is (\overline{ab.cd}), where all digits are nonzero. The same number in base twelve is (\overline{bb.ba}). Find


[

a+b+c.

]


---


5. Cut off a square carton by a straight line into two pieces, then cut one of two pieces into two small pieces by a straight line, etc. By cutting 2017 times we obtain 2018 pieces. We write number 2 in every triangle, number 1 in every quadrilateral, and 0 in the polygons. Is the sum of all inserted numbers always greater than 2017?

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